If the curve $y=x^2+bx+c$ touches the line $y=x$ at the point (1, 1) then the values of x for which the curve has a negative gradient are.      

$y=x^2+bx+c$

$\frac{dy}{dx}=2x+b=2x+b=1$

$\frac{dy}{dx}/\left(1,1\right)=2+b=1=b=-1$$y=x^2+bx+c$

$y=x^2+bx+c$

$1=1+b+c$

$c=-b$

$c=1$

$\therefore y=x^2-x+1\Rightarrow \frac{dy}{dx}=2x-1$

$\therefore$ curve has a negative gradient 

$m<0$

$2x-1<0$

$2x<1$

$x<1/2$